Given an array of integers, the task is to find the minimum number of segments that the array elements can be divided into such that all the segments contain distinct elements.
Examples:
Input: n = 6; Array: 1, 7, 4, 3, 3, 8
Output: 2
Explanation:
Optimal way to create segments here is {1, 7, 4, 3} {3, 8}
Clearly, the answer is the maximum frequency of any element within the array i.e. '2'.
as '3' is the element which appears the most in the array (twice).
Input : n = 6; Array: 2, 2, 3, 3, 3, 5
Output : 4
Below is the implementation of the above approach:
- Count the number of segments required
- For counting the frequency
- Iterate over the array
- Increment the frequency
- Check if there is any duplicate in current segment.
- Increment the segment required by 1.
- Clear the map
- Increment the frequency of current element in map.
- Return the count.
C++
#include <bits/stdc++.h>using namespace std;// Function to count the minimum minimum number of segments// that the array elements can be divided into such that all// the segments contain distinct elements.int countSegment(vector<int>& arr){ int n = arr.size(); // count the number of segment required int count = 1; // For counting the frequency unordered_map<int, int> unmap; // Iterate over the array for (int i = 0; i < n; i++) { // Increment the frequency unmap[arr[i]]++; // Check if there is any duplicate in current // segment. if (unmap[arr[i]] > 1) { // Increment the segment required by 1. count++; // Clear the map unmap.clear(); // Increment the frequency of current element in // map unmap[arr[i]]++; } } // Return the count. return count;}int main(){ // Input array. vector<int> arr = { 2, 2, 3, 3, 3, 5 }; // Function call and store the result int result = countSegment(arr); // Print the result cout << result << endl; return 0;} |
Java
import java.util.*;public class Main { // Function to count the minimum minimum number of // segments that the array elements can be divided into // such that all the segments contain distinct elements. public static int countSegment(ArrayList<Integer> arr) { int n = arr.size(); // count the number of segments required int count = 1; // For counting the frequency HashMap<Integer, Integer> map = new HashMap<Integer, Integer>(); // Iterate over the array for (int i = 0; i < n; i++) { // Increment the frequency map.put(arr.get(i), map.getOrDefault(arr.get(i), 0) + 1); // Check if there is any duplicate in current // segment. if (map.get(arr.get(i)) > 1) { // Increment the segment required by 1. count++; // Clear the map map.clear(); // Increment the frequency of current // element in map map.put(arr.get(i), 1); } } // Return the count. return count; } public static void main(String[] args) { // Input array. ArrayList<Integer> arr = new ArrayList<Integer>(); arr.add(2); arr.add(2); arr.add(3); arr.add(3); arr.add(3); arr.add(5); // Function call and store the result int result = countSegment(arr); // Print the result System.out.println(result); }} |
Python3
# Function to count the minimum number of segments that the array elements# can be divided into such that all the segments contain distinct elements.def countSegment(arr): n = len(arr) # Count the number of segments required count = 1 # For counting the frequency unmap = {} # Iterate over the array for i in range(n): # Increment the frequency if arr[i] not in unmap: unmap[arr[i]] = 1 else: unmap[arr[i]] += 1 # Check if there is any duplicate in current segment. if unmap[arr[i]] > 1: # Increment the segment required by 1. count += 1 # Clear the map unmap = {} # Increment the frequency of current element in map unmap[arr[i]] = 1 # Return the count. return count# Input arrayarr = [2, 2, 3, 3, 3, 5]# Function call and store the resultresult = countSegment(arr)# Print the resultprint(result) |
C#
using System;using System.Collections.Generic;class Program { // Function to count the minimum minimum number of // segments that the array elements can be divided into // such that all the segments contain distinct elements. static int CountSegment(List<int> arr) { int n = arr.Count; // count the number of segment required int count = 1; // For counting the frequency Dictionary<int, int> dict = new Dictionary<int, int>(); // Iterate over the array for (int i = 0; i < n; i++) { // Increment the frequency if (dict.ContainsKey(arr[i])) dict[arr[i]]++; else dict.Add(arr[i], 1); // Check if there is any duplicate in current // segment. if (dict[arr[i]] > 1) { // Increment the segment required by 1. count++; // Clear the map dict.Clear(); // Increment the frequency of current // element in map dict[arr[i]] = 1; } } // Return the count. return count; } static void Main(string[] args) { // Input array. List<int> arr = new List<int>{ 2, 2, 3, 3, 3, 5 }; // Function call and store the result int result = CountSegment(arr); // Print the result Console.WriteLine(result); }} |
Javascript
// Function to count the minimum number of segments that the array elements can be divided into such that all the segments contain distinct elements.function countSegment(arr) { const n = arr.length; // count the number of segment required let count = 1; // For counting the frequency const unmap = new Map(); // Iterate over the array for (let i = 0; i < n; i++) { // Increment the frequency if (unmap.has(arr[i])) { unmap.set(arr[i], unmap.get(arr[i]) + 1); } else { unmap.set(arr[i], 1); } // Check if there is any duplicate in current segment. if (unmap.get(arr[i]) > 1) { // Increment the segment required by 1. count++; // Clear the map unmap.clear(); // Increment the frequency of current element in map unmap.set(arr[i], 1); } } // Return the count. return count;}// Input array.const arr = [2, 2, 3, 3, 3, 5];// Function call and store the resultconst result = countSegment(arr);// Print the resultconsole.log(result); |
Output
4
Time Complexity: O(n)
Auxiliary Space: O(n)
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